A problem of configurations: a(0) = 1; for n>0, a(n) = (2n-1)!! - Sum_{k=1..n-1} (2k-1)!! a(n-k). Also the number of shellings of an n-cube, divided by 2^n n!.
1, 1, 2, 10, 74, 706, 8162, 110410, 1708394, 29752066, 576037442, 12277827850
1 seqfan posts
Wed Dec 29 15:36:54 CET 2010 [seqfan] A000698 == indecomposable fixed-point free involutions of {2n}?